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Diberikan fungsi f(x)=x+4​a−2​ dan g(x)=2bx+5−6x−5​. Jika 2(f⋅g)(5)=−3(gf​)(0) dan −f(12)+g(1)=0, maka nilai dari 10ab+21a−20b adalah ....

Pertanyaan

Diberikan fungsi f left parenthesis x right parenthesis equals fraction numerator a minus 2 over denominator square root of x plus 4 end root end fraction dan g left parenthesis x right parenthesis equals fraction numerator negative 6 x minus 5 over denominator 2 b x plus 5 end fraction. Jika 2 left parenthesis f times g right parenthesis left parenthesis 5 right parenthesis equals negative 3 open parentheses f over g close parentheses left parenthesis 0 right parenthesis dan negative f left parenthesis 12 right parenthesis plus g left parenthesis 1 right parenthesis equals 0, maka nilai dari 10 a b plus 21 a minus 20 b adalah ....

  1. begin mathsize 14px style negative 20 end style

  2. begin mathsize 14px style negative 10 end style

  3. begin mathsize 14px style 10 end style

  4. begin mathsize 14px style 20 end style

  5. begin mathsize 14px style 40 end style

D. Natalia

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah D.

Pembahasan

Ingat kembali bahwa begin mathsize 14px style left parenthesis f times g right parenthesis left parenthesis x right parenthesis equals f left parenthesis x right parenthesis times g left parenthesis x right parenthesis end style.

Oleh karena itu, didapat perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis f times g right parenthesis left parenthesis 5 right parenthesis end cell equals cell f left parenthesis 5 right parenthesis times g left parenthesis 5 right parenthesis end cell row blank equals cell open parentheses fraction numerator a minus 2 over denominator square root of 5 plus 4 end root end fraction close parentheses times open parentheses fraction numerator negative 6 times 5 minus 5 over denominator 2 b times 5 plus 5 end fraction close parentheses end cell row blank equals cell open parentheses fraction numerator a minus 2 over denominator square root of 9 end fraction close parentheses times open parentheses fraction numerator negative 30 minus 5 over denominator 5 open parentheses 2 b plus 1 close parentheses end fraction close parentheses end cell row blank equals cell open parentheses fraction numerator a minus 2 over denominator 3 end fraction close parentheses times open parentheses fraction numerator negative 35 over denominator 5 open parentheses 2 b plus 1 close parentheses end fraction close parentheses end cell row blank equals cell open parentheses fraction numerator a minus 2 over denominator 3 end fraction close parentheses times open parentheses fraction numerator negative 7 over denominator 2 b plus 1 end fraction close parentheses end cell row blank equals cell negative fraction numerator 7 open parentheses a minus 2 close parentheses over denominator 3 open parentheses 2 b plus 1 close parentheses end fraction end cell row blank equals cell fraction numerator negative 7 a plus 14 over denominator 6 b plus 3 end fraction space... space open parentheses straight i close parentheses end cell end table


Kemudian, ingat pula bahwa begin mathsize 14px style open parentheses f over g close parentheses left parenthesis x right parenthesis equals fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction comma space g left parenthesis x right parenthesis not equal to 0 end style.

Oleh karena itu, didapat perhitungan sebagai berikut. 

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f over g close parentheses left parenthesis 0 right parenthesis end cell equals cell fraction numerator f left parenthesis 0 right parenthesis over denominator g left parenthesis 0 right parenthesis end fraction end cell row blank equals cell fraction numerator open parentheses begin display style fraction numerator a minus 2 over denominator square root of 0 plus 4 end root end fraction end style close parentheses over denominator open parentheses begin display style fraction numerator negative 6 times 0 minus 5 over denominator 2 b times 0 plus 5 end fraction end style close parentheses end fraction end cell row blank equals cell fraction numerator open parentheses begin display style fraction numerator a minus 2 over denominator square root of 4 end fraction end style close parentheses over denominator open parentheses begin display style fraction numerator negative 5 over denominator 5 end fraction end style close parentheses end fraction end cell row blank equals cell fraction numerator open parentheses begin display style fraction numerator a minus 2 over denominator 2 end fraction end style close parentheses over denominator open parentheses begin display style negative 1 end style close parentheses end fraction end cell row blank equals cell negative fraction numerator a minus 2 over denominator 2 end fraction space... space left parenthesis ii right parenthesis end cell end table


Subtitusikan persamaan (i) dan (ii) ke persamaan size 14px 2 begin mathsize 14px style left parenthesis f times g right parenthesis end style begin mathsize 14px style left parenthesis 5 right parenthesis end style size 14px equals size 14px minus size 14px 3 begin mathsize 14px style open parentheses f over g close parentheses end style begin mathsize 14px style left parenthesis 0 right parenthesis end style sehingga diperoleh perhitungan sebagai berikut.

table row cell 2 open parentheses f times g close parentheses open parentheses 5 close parentheses equals negative 3 open parentheses f over g close parentheses open parentheses 0 close parentheses end cell row cell 2 open parentheses fraction numerator negative 7 a plus 14 over denominator 6 b plus 3 end fraction close parentheses equals negative 3 open parentheses negative fraction numerator a minus 2 over denominator 2 end fraction close parentheses end cell row cell fraction numerator negative 14 a plus 28 over denominator 6 b plus 3 end fraction equals fraction numerator 3 a minus 6 over denominator 2 end fraction end cell row cell 2 left parenthesis negative 14 a plus 28 right parenthesis equals left parenthesis 6 b plus 3 right parenthesis left parenthesis 3 a minus 6 right parenthesis end cell row cell negative 28 a plus 56 equals 18 a b minus 36 b plus 9 a minus 18 end cell row cell negative 18 a b minus 37 a plus 36 b equals negative 74 end cell row cell 18 a b plus 37 a minus 36 b equals 74 space... space left parenthesis iii right parenthesis end cell end table


Dari soal diketahui bahwa begin mathsize 14px style negative f left parenthesis 12 right parenthesis plus g left parenthesis 1 right parenthesis equals 0 end style.

Oleh karena itu, didapat perhitungan sebagai berikut.

table row cell negative f left parenthesis 12 right parenthesis plus g left parenthesis 1 right parenthesis equals 0 end cell row cell negative fraction numerator a minus 2 over denominator square root of 12 plus 4 end root end fraction plus fraction numerator negative 6 times 1 minus 5 over denominator 2 b times 1 plus 5 end fraction equals 0 end cell row cell negative fraction numerator a minus 2 over denominator square root of 16 end fraction plus fraction numerator negative 6 minus 5 over denominator 2 b plus 5 end fraction equals 0 end cell row cell negative fraction numerator a minus 2 over denominator 4 end fraction minus fraction numerator 11 over denominator 2 b plus 5 end fraction equals 0 end cell row cell negative fraction numerator a minus 2 over denominator 4 end fraction equals fraction numerator 11 over denominator 2 b plus 5 end fraction end cell row cell negative left parenthesis a minus 2 right parenthesis left parenthesis 2 b plus 5 right parenthesis equals 11 times 4 end cell row cell left parenthesis a minus 2 right parenthesis left parenthesis 2 b plus 5 right parenthesis equals negative 44 end cell row cell 2 a b plus 5 a minus 4 b minus 10 equals negative 44 end cell row cell 2 a b plus 5 a minus 4 b equals negative 34 space... space left parenthesis iv right parenthesis end cell end table

Jumlahkan persamaan (iii) dan (iv) sehingga diperoleh hasil sebagai berikut.

table attributes columnalign right right right right columnspacing 2px 2px 2px 0.4em end attributes row cell 18 a b plus end cell cell 37 a minus end cell cell 36 b equals end cell 74 space row cell 2 a b plus end cell cell 5 a minus end cell cell 4 b equals end cell cell negative 34 end cell plus row cell 20 a b plus end cell cell 42 a minus end cell cell 40 b equals end cell 40 space row cell 10 a b plus end cell cell 21 a minus end cell cell 20 b equals end cell 20 space end table

Dengan demikian, nilai dari begin mathsize 14px style 10 a b plus 21 a minus 20 b end style adalah undefined.

Jadi, jawaban yang tepat adalah D.

52

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Pertanyaan serupa

Perhatikan gambar berikut ini!   Nilai dari (f ∙ g) (3)  dan domain fungsi (f ∙ g) (x)  adalah ....

13

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